Question

Titration problem?

A weak acid HA is titrated with a standard base ( NaOH solution) and the following titration curve is obtained.

For a weak acid , the dissociation constant , Ka , is defined as:

HA + H2O <=> `A^- + H3O^+`

Ka= [A-][H3O+]/[HA]

The realtionship between Ka and pH is :

-lgKa=pH - lg{[A-]/[HA]}

and pH = -lg[H3O+]

Using the information provided determine the dissociation constant ,Ka , for the weak acid HA.

a) `10^{-4,2}`

b)`10^{-4,5}`

c)`10^{-7,8}`

d)`10^{-11,5}`

Answer 1

`"Ka"= 10^(-4.5)`

Explanation:

The `"pKa"` of a monoprotic weak acid is the same as the `"pH"` halfway to its equivalence point. Here's how this shortcut works:

Sodium hydroxide `"NaOH"` neutralizes the weak acid `"HA"` to produce `"NaA"` by the equation

`"NaOH" + "HA" to "NaA" + "H"_2"O" color(grey)(" balanced.")`

One formula unit of sodium hydroxide reacts with one `"HA"` molecule to produce one formula unit of `"NaA"`. Thus

`n("NaA")=n("HA","reacted")=n("NaOH", "added")`

Appearantly
`n("HA", "reacted")+ n("HA", "remaining") = n("HA", "initial")`

Substituting `n("HA","reacted")` in the second equation with `n("NaA")` gives

`n("HA","remaining") = n("HA", "initial")- n("NaA")`

`n("NaA") = n("NaOH") = n("HA", "initial")` at the equivalence point where `V("NaOH", "added") = 22 color(white)(l) "ml"` for this question;

Therefore halfway to that point at the titration midpoint at which `11 color(white)(l) "ml"` of `"NaOH"` has been added, `n("NaA") = n("NaOH") = color(navy)(1/2)n("HA", "initial")`

`n("HA","remaining") = n("HA", "initial")- n("NaA")`
`color(white)(n("HA","remaining")) =color(navy)(1/2)n("HA", "initial")`
`color(white)(n("HA","remaining") )=n("NaA")`

Additionally, given that

`"NaA" (aq) color(purple) to "Na"^(+) + "A"^-`

`"NaA"`, a salt, completely disassociates as a strong electrolyte,

`n("A"^-)=n("NaA")= n("HA")`.

`"Ka"` is defined as the equilibrium constant of the reaction

`"HA" (aq) rightleftharpoons "H"^+ (aq) + "A"^(-) (aq)`

`"Ka"="K"_"eq"=(["H"^+] color(red)(cancel(color(black)(["A"^(-)]))))/(color(red)(cancel(color(black)(["HA"]))))=["H"^+]`

given that `n("A"^-)=n("NaA")= n("HA")` as previously stated.

Therefore

`"Ka"=["H"^+]=10^(-"pH")=10^(-4.5)`

Reference
"Titration Fundamentals", Chemistry LibreText, https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_Fundamentals

See also
"Titration of a Weak Base with a Strong Acid", https://chem.libretexts.org/Demos%2C_Techniques%2C_and_Experiments/General_Lab_Techniques/Titration/Titration_of_a_Weak_Base_with_a_Strong_Acid

Answer 2

`K_a = 10^(-4.5)`

Explanation:

As an alternative approach, you can use the fact that at the half-equivalence point, exactly half of the weak acid has been neutralized by the strong base.

The chemical equation that describes this neutralization reaction looks like this

`"HA"_ ((aq)) + "OH"_ ((aq))^(-) -> "A"_ ((aq))^(-) + "H"_ 2"O"_ ((l))`

At the half-equivalence point, you add enough moles of strong base, which I've represented here by `"OH"^(-)`, to consume half of the initial number of moles of the weak acid.

Since the reaction consumes the weak acid and produces the conjugate base `"A"^(-)` in a `1:1` mole ratio, you can say that at the half-equivalence point, you have

`["HA"] = ["A"^(-)]`

Now, notice that at the equivalence point, `"22 mL"` of the strong base have been added to the weak acid solution. This tells you that it takes `"22 mL"` of the strong base to consume all the moles of the weak acid present in the initial solution.

So if it takes `"22 mL"` of the strong base to consume all the moles of the weak acid, it follows that if you add `"11 mL"` of the strong base, you will consume exactly half of the moles of the weak acid.

This means that the half-equivalence point corresponds to `"11 mL"` of the strong base added and a `"pH"` of `4.5`.

The question provides you with the equation

`-"lg"(K_a) = "pH" - "lg"( (["A"^(-)])/(["HA"]))`

which is a version of the Henderson - Hasselbalch equation that can be used to calculate the `"pH"` of a weak acid - conjugate base buffer.

Plug in the `"pH"` of the solution at the half-equivalence point and the fact that `["HA"] = ["A"^(-)]` to get

`- "lg"(K_a)= 4.5 - "lg"(1)`

This gets you

`"lg"(K_a) = - 4.5`

which implies

`color(darkgreen)(ul(color(black)(K_a = 10^(-4.5))))`

See this answer for a similar question.

SIDE NOTE: The notation used here is based on the fact that

`"lg"(x) = log_(10)(x)`

See here for more info on that.